Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Welcome To Ask or Share your Answers For Others

Categories

0 votes
141 views
in Technique[技术] by (71.8m points)

Exposing a C++ API to Python

I'm currently working on a project were I had to wrap the C++ classes with Python to be able to script the program. So my specific experience also involved embedding the Python interpreter in our program.

The alternatives I tried were:

  • Boost.Python

    I liked the cleaner API produced by Boost.Python, but the fact that it would have required that users install an additional dependency made us switch to SWIG.

  • SWIG

    SWIG's main advantage for us was that it doesn't require end users to install it to use the final program.

What have you used to do this, and what has been your experience with it?

question from:https://stackoverflow.com/questions/276761/exposing-a-c-api-to-python

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
Welcome To Ask or Share your Answers For Others

1 Reply

0 votes
by (71.8m points)

I've used both (for the same project): Boost is better integrated with the STL, and especially C++ exceptions. Also, its memory management mechanism (which tries to bridge C++ memory management and Python GC) is way more flexible than SWIG's. However, SWIG has much better documentation, no external dependencies, and if you get the library wrapped in SWIG for Python you're more than half-way there to getting a Java/Perl/Ruby wrapper as well.

I don't think there's a clear-cut choice: for smaller projects, I'd go with Boost.Python again, for larger long-lived projects, the extra investment in SWIG is worth it.


与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
OGeek|极客中国-欢迎来到极客的世界,一个免费开放的程序员编程交流平台!开放,进步,分享!让技术改变生活,让极客改变未来! Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Click Here to Ask a Question

...